codeforces 385C 题目大意: 给出n个数,数据范围107,给出m次查询,每次给出l,r求在l和r之间的素数的∑ki=0f(pi),f(pi)定义为n个数中能够整除pi的个数。 题目分析: 直接线性筛素数,记录每个数最大的素因数,然后统计对于每个素因数对应的f(pi),然后求取前缀和,来方便求取区间和。 AC代码: #include <...
385C - Bear and Prime Numbers 思路:记录数组中1-1e7中每个数出现的次数,然后用素数筛看哪些能被素数整除,并加到记录该素数的数组中,然后1-1e7求一遍前缀和。 代码: #include<bits/stdc++.h>usingnamespacestd;#definell long long#definepb push_back#definemem(a,b) memset((a),(b),sizeof(a))co...
CodeForces - 385C Bear and Prime Numbers——素数筛法 在筛素数时计算每个数可以对应多少个题目中给定的数ai, 满足这个数是ai的因子, 然后预处理前缀和,O(1)完成查询 #include#include#include#includeusing namespace std;const int maxn = 1e6;const int maxm = 1e7 + 5;int n, m, a[maxn], l,...
Secondary School MathematicsA Computational Perspectivedoi:10.1007/0-387-28979-8Richard CrandallCenter for Advanced Computation, 3754 SE Knight Street, Portland, OR, 97202, USACarl PomeranceDepartment of Mathematics, Dartmouth College, Hanover, NH, 03755-3551, USASpringer New York...
prime numbers defined Prime numbers are integers greater than one that are only evenly divisible (no remainder) by one and themselves. That is, their only integer factors are one (1) and the number itself. rime numbers have no integer divsors other than one and themselves. yes, I have hin...
cf C Bear and Prime Numbers,题意:给你一个n,输入n个数,然后输入m,接下来有m个询问,每一个询问为[l,r],然后输出在区间内[l,r]内f(p)的和,p为[l,r]的素数,f(p)的含义为在n个数中是p的倍数的个数。思路:先打出10000000内的素数,然后统计每一个素数在n个数中的倍数
题目链接:Codeforces 385C Bear and Prime Numbers 这题告诉我仅仅有询问没有更新通常是不用线段树的。或者说还有比线段树更简单的方法。 用一个sum数组记录前n项和,这个sum数组在打素数表时候就能够求出来,注意一点求素数的内层循环要改成i。不能再写成i + i或者i * i了。原因想想就明确了。
primesieve is a command-line program and C/C++ library for quickly generating prime numbers. It is very cache efficient, it detects your CPU's L1 & L2 cache sizes and allocates its main data structures accordingly. It is also multi-threaded by default, it uses all available CPU cores ...
CheckPrimeNumbers importjava.io.*; publicclassCheckPrimeNumbers{ publicstaticvoidmain(String[]args)throwsException{ intcount=0;//Countthenumberofprimenumbers DataInputStreaminput= newDataInputStream(newBufferedInputStream( newFileInputStream(args[0]))); long[]limits={ 10, 100, 1000, 10000, 1...
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